[leetcode]20.Valid Parentheses

题目

Given a string containing just the characters '(', ')', '{', '}', '[' and ']', determine if the input string is valid.

An input string is valid if:

Open brackets must be closed by the same type of brackets.
Open brackets must be closed in the correct order.
Note that an empty string is also considered valid.

Example 1:

Input: "()"
Output: true
Example 2:

Input: "()[]{}"
Output: true
Example 3:

Input: "(]"
Output: false
Example 4:

Input: "([)]"
Output: false
Example 5:

Input: "{[]}"
Output: true

解法

思路

是个很简单的括号匹配问题，用栈来解决。碰到左括号入栈。碰到右括号，若此时栈为空，则直接返回false，如果不为空，则与栈顶括号对比，如果匹配成功则栈顶元素出栈，然后继续循环。如果匹配失败，直接返回false。等所有括号都扫描完成后，如果栈内为空，则返回true，否则返回false。

代码

class Solution {
    public boolean isValid(String s) {
        Stack<Character> stack = new Stack<Character>();
        for(int i = 0; i < s.length();i++){
            if(s.charAt(i) == '(' || s.charAt(i) == '[' || s.charAt(i) == '{')
                stack.push(s.charAt(i));
            else{
                if(stack.isEmpty()) return false;
                if(s.charAt(i) == ')'&&stack.pop()!='(') return false;
                if(s.charAt(i) == ']'&&stack.pop()!='[') return false;
                if(s.charAt(i) == '}'&&stack.pop()!='{') return false;
            }   
        }
        return stack.isEmpty();
    }
}

更优雅的代码（原理相同）

class Solution {
    public boolean isValid(String s) {
        Stack<Character> stack = new Stack<Character>();
        for(char c:s.toCharArray()){
            if(c == '(')
                stack.push(')');
            else if(c == '[')
                stack.push(']');
            else if(c == '{')
                stack.push('}');
            else if(stack.isEmpty() || stack.pop()!=c)
                return false;
        }
        return stack.isEmpty();
    }
}